Starting from home (x=0,y=0) you make thre successive displacements, resting aft

Question

Starting from home (x=0,y=0) you make thre successive displacements, resting after eachmovement. First you go 5m due West, followed by a motion withacceleration of 2 m/s^2 for 2s 20 degrees East of North.Finally you go 15m, 10degrees South of East. a.) find each displacement vector in the unit vector notation.East direction is characterized by i^ and north direction byj^. b.) find the net displacement vector. c.) find the magnitude of the displacement vector and in whichquadrant it is in. Starting from home (x=0,y=0) you make thre successive displacements, resting after eachmovement. First you go 5m due West, followed by a motion withacceleration of 2 m/s^2 for 2s 20 degrees East of North.Finally you go 15m, 10degrees South of East. a.) find each displacement vector in the unit vector notation.East direction is characterized by i^ and north direction byj^. b.) find the net displacement vector. c.) find the magnitude of the displacement vector and in whichquadrant it is in.

Explanation / Answer

You start at (0,0) then you go 5 m due west, which is thedirection -x, so you are now at (-5,0). Then you accelerate at 2 m/s2 for 2 seconds. So your final velocity is 4 m/s. And your average velocity is(4+0)/2 = 2 m/s. You did this for 2 seconds, so you went 4meters. Draw a triangle 20 degrees to the right of thevertical. You should notice that the change in x will be 4m *sin(20), and the change in y will be 4m * cos(20). So addthis to your second point, and you will be at: (-3.6319, 3.7588). Then you move 15 m, 10 degrees southof east. Again draw a triangle and you will see that yourchange in x is 15 * cos(10) and your change in y is 15 *sin(10). So your final position is (11.1402, 1.154). In vector notation, this is 11.1402 x^ + 1.154 y^ You can get the length of the vector with the pothagoreanformula sqrt(11.1402^2 + 1.154^2)=11.2 And you can get your angle using tan-1(1.154/11.1402)=5.9 degrees north of east So your net displacement vector is 11.2, 5.9 degrees north ofeast And the magnitude of the displacement vector is 11.2. Itis in the 1st quadrant since x and y are both positive.

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